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A cyclic group is a group generated by a single element. In other words, given some "generator" g in G, every element in G can be represented as a power of g. Since G and Z_n are both cyclic groups of order n, we can write: \( G = \{g^0, g^1, g^2,..., g^{n-1}\} \) and \(Z_n = \{1, \omega, \omega^2,..., \omega^{n-1}\}\), where g and ω are generators of G and Z_n, respectively.

A homomorphism is a map between two groups that preserves the group structure. Given the cyclic groups G and Z_n, we can define a homomorphism, φ, as follows: \(φ: G \rightarrow Z_n\), such that \(φ(g^k)=\omega^{ak}\), where "a" is a constant factor. The role of a homomorphism is to preserve group operations, so we have \(φ(g^k) * φ(g^l) = φ(g^{k+l})\).

Now we need to check the properties of the homomorphism φ: 1. Identity: The identity element of G is \(g^0\), and the identity element of Z_n is 1. We have \(φ(g^0) = \omega^{a*0} = 1\), so the identity element is preserved. 2. Inverses: For any element \(g^k\) in G, its inverse is \(g^{-k}\). The inverse of \(φ(g^k)=\omega^{ak}\) in Z_n is \(\omega^{-ak}\). We have: \( φ(g^{-k}) = \omega^{-ak} = (\omega^{ak})^{-1} \), so the inverses are preserved. 3. Group operation preservation: For any two elements \(g^k, g^l \in G\), their product in G is \(g^{k+l}\). We have: \(φ(g^k) * φ(g^l) = \omega^{ak} * \omega^{al} = \omega^{a(k+l)} = φ(g^{k+l})\), so the group operation is preserved. Thus, φ is a homomorphism.

Let H be the group of all homomorphisms from G to Z_n, H = {φ_0, φ_1, ..., φ_(n-1)}. Each homomorphism φ_i is defined by a different constant factor "a" (i.e., φ_i(g^k) = \(\omega^{i*k}\)). We will show that H is cyclic by proving that it has a generator. Consider the homomorphism \(φ_{1}\) with constant factor a = 1. We claim that \(φ_{1}\) generates H. To prove this, we need to show that every homomorphism in H can be expressed as a power of \(φ_{1}\). For each φ_i, where 0 ≤ i ≤ n-1, we have: \( (φ_{1}^i)(g^k) = φ_{1}(g^{ik}) = (\omega^k)^i = \omega^{i*k} = φ_i(g^k) \) This shows that each homomorphism φ_i can be represented as a power of \(φ_{1}\), so H is cyclic with generator \(φ_{1}\).

Now we need to find the order of the cyclic group H, generated by \(φ_{1}\). Note that H has n elements, since there are n different possible constant factors "a" for the homomorphisms. This means the order of H is n. In conclusion, the group of homomorphisms from G to Z_n is a cyclic group of order n, generated by \(φ_{1}\).